Integration

\hspace{-16}$I Dont have much Idea about that type of Integral....\\\\\\ May be it is an Elleptical Integral of $\bf{(I)}$ or $\bf{(II)}$ Type..\\\\\\Given $\bf{\int\frac{x^2}{\sqrt{1-x^4}}\;dx = \int x^2\cdot \left(1-x^4\right)^{-\frac{1}{2}}dx}$\\\\\\ Now Using Chebyshev integral....\\\\\\ $\bf{\bullet \;\; \int x^m\cdot (a+bx^n)^{p}\; dx}$\\\\\\ Where $\bf{m,n,p}$ are Rational no. and the given Integral\\\\\\ is Integral in elementry form only if ............\\\\\\ $\bf{\left(\frac{m+1}{n}\right)\in \mathbb{Z}}$ \bf{\underline{\underline{or}}} $\bf{\left(\frac{m+1}{n}+p\right)\in \mathbb{Z}}$\\\\\\ Now in $\bf{\int x^2\cdot \left(1-x^4\right)^{-\frac{1}{2}}dx}$\\\\\\ Here $\bf{m=2\;\;,a=1\;\;,b=-1\;\;,n=4\;\;,p=-\frac{1}{2}}$\\\\\\

\hspace{-20}$So $\bf{\left(\frac{2+1}{4}\right)\neq \mathbb{Z}}$ \bf{\underline{\underline{or}}} $\bf{\left\{\left(\frac{2+1}{4}-\frac{1}{2}\right)\right\}\neq \mathbb{Z}}$\\\\\\ So we can Not Integrate $\bf{\int\frac{x^2}{\sqrt{1-x^4}}\;dx}$ in terms of\\\\\\ elementry function..